Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.Arrays;
import java.util.StringTokenizer;


public class Solution {
    // VM options: -Xmx1024m -Xss1024m

    private static String inputFilename = "input.txt";
    private static String outputFilename = "output.txt";
    private BufferedReader in;
    private StringTokenizer line;
    private PrintWriter out;
    private boolean isDebug;

    public Solution(boolean isDebug) {
        this.isDebug = isDebug;
    }


    public void solve() throws IOException {
        int n = nextInt();
        int m = nextInt();
        int[][] a = new int[n][];
        int mni = 0;
        int mnj = 0;
        for (int i = 0; i < n; i++) {
            a[i] = nextIntArray(m);
            for (int j = 0; j < m; j++) {
                if (a[mni][mnj] > a[i][j]) {
                    mni = i;
                    mnj = j;
                }
            }
        }
        int[] row = new int[m];
        int[] col = new int[n];
        int res = 1000000000;
        int resk = -1;
        for (int k = 0; k <= a[mni][mnj]; k++) {
            int tr = 0;
            for (int i = 0; i < n; i++) {
                col[i] = a[i][mnj] - k;
                tr += col[i];
            }
            for (int j = 0; j < m; j++) {
                row[j] = a[mni][j] - (a[mni][mnj] - k);
                tr += row[j];
            }
            if (tr < res) {
                boolean pp = true;
                for (int i = 0; i < n && pp; i++) {
                    for (int j = 0; j < m; j++) {
                        if (i != mni && j != mnj && a[i][j] != row[j] + col[i]) {
                            pp = false;
                            break;
                        }
                    }
                }
                if (pp) {
                    res = tr;
                    resk = k;
                }
            }
        }
        if (resk < 0) {
            out.println(-1);
        } else {
            out.println(res);
            for (int i = 0; i < resk; i++) {
                out.println("col " + (mnj + 1));
            }
            for (int i = 0; i < a[mni][mnj] - resk; i++) {
                out.println("row " + (mni + 1));
            }
            for (int i = 0; i < n; i++) {
                if (i != mni) {
                    for (int x = 0; x < a[i][mnj] - resk; x++) {
                        out.println("row " + (i + 1));
                    }
                }
            }
            for (int j = 0; j < m; j++) {
                if (j != mnj) {
                    for (int x = 0; x < a[mni][j] - (a[mni][mnj] - resk); x++) {
                        out.println("col " + (j + 1));
                    }
                }
            }
        }
    }

    public static void main(String[] args) throws IOException {
        new Solution(Arrays.asList(args).contains("DEBUG_MODE")).run(args);
    }

    public void run(String[] args) throws IOException {
        if (isDebug) {
            in = new BufferedReader(new InputStreamReader(new FileInputStream(inputFilename)));
//            in = new BufferedReader(new InputStreamReader(System.in));
        } else {
            in = new BufferedReader(new InputStreamReader(System.in));
        }
        out = new PrintWriter(System.out);
//        out = new PrintWriter(outputFilename);

//        int t = nextInt();
        int t = 1;
        for (int i = 0; i < t; i++) {
//            out.print("Case #" + (i + 1) + ": ");
            solve();
        }

        in.close();
        out.flush();
        out.close();
    }

    private int[] nextIntArray(int n) throws IOException {
        int[] res = new int[n];
        for (int i = 0; i < n; i++) {
            res[i] = nextInt();
        }
        return res;
    }

    private long[] nextLongArray(int n) throws IOException {
        long[] res = new long[n];
        for (int i = 0; i < n; i++) {
            res[i] = nextLong();
        }
        return res;
    }

    private int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }

    private long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }

    private double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }

    private String nextToken() throws IOException {
        while (line == null || !line.hasMoreTokens()) {
            line = new StringTokenizer(in.readLine());
        }
        return line.nextToken();
    }
}